Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 189

New comment
 
Mathemat:

Another one:

There are 2,000 balls that look the same, half of which are aluminium and half are dural. Balls of the same material weigh the same, balls of different materials weigh differently. What minimum number of weighings on a cup scale will be needed to ensure the formation of two groups of different weight from the same number of balls?

The weight is 4.

FAQ:

- The scales are cup scales, infinitely accurate, there are no weights. Weighing is putting something on both bowls, looking at the balance, remembering the result and removing the contents from the bowls,

- Wiki says that the density of dural is about equal to that of aluminium. For this problem, it is sufficient to assume that it is simply different from the density of aluminium,

- formed groups of different weights of the same number of balls can have any number of balls, even one at a time,

- proving the minimum number of weights is necessary - unless, of course, you have managed for the minimum possible number of weights.

1999

ZS: even probably 1998

 
no, you can even do it for 1,000 weighings )
 
for 4, I think it's possible.
 
TheXpert:
for 4 I think you can.
Not for 4 you can't and for 1,000 you can't, guaranteed.
 

you need to apply array sorting techniques here,

which one is the most economical?

I can't do it in less than 1998.

 

Although if the task is not to divide into aluminium and dural, no weighing is needed at all, we divide into two parts with the same number of balls,

and the weight will be different for sure.

 
sanyooooook:
You have begun to understand ) not for sure
 
so wait, the task of splitting the whole group of balls into two is not an issue, you can just form 2 piles of 2 balls each?
 
sanyooooook:
so wait, the task is not to divide the whole group of balls into two, can we just form 2 piles of 2 balls each?

And, if not all of them need to be separated, then one weighing )

If all of them, then two, not 4.

 
TheXpert:

And if you don't want to divide them all, then one weighing)


What if there are two of them at once?

) We take out the third one, but what if it is the same as the first one? )

the fourth, but what if it's the same as the first?

the fifth? that one would weigh the same as the first.)

The probability of such an event is small, but it exists,

which means there's no guarantee it'll happen.

1...182183184185186187188189190191192193194195196...229
New comment